Solution

According to the problem statement, it is required to find the minimum mass of the particles in the beam at which all of them will reach the second grid. In this case, the work of the force between the grid electrodes over the distance $\ell$ is equal to the change in the kinetic energy of the particles:
$$F\ell = \Delta E_k = \frac{mv^2}{2},$$
here it is taken into account that the work of the force $F$ is negative, and the change in kinetic energy is:
$$\Delta E_k = 0 - \frac{mv^2}{2}$$

Then the minimum mass of the particles is:
$$m = \frac{2F\ell}{v^2}$$

Note: the problem can also be solved dynamically, using the impulse of force. Try to reach the answer independently using this method.